Principle deep dive
Modulus / absolute value behaviour
Splitting |·| at its zero. The principle behind the 2023 modulus spike — the broadest cross-chapter reach in the bank at 11 chapters, from limits and derivatives through definite integrals, areas, equations and relations. Includes the disguised form, where √(f²) is really |f|.
- questions in the bank
- 94
- tagged HARD
- 14%
- chapter spread
- 11
- worked examples below
- 4
When to reach for it
The expression contains |·| — or a square root of something squared, which is |·| in disguise.
Why this principle matters
|x| splits at zero: equals x for x ≥ 0, equals −x for x < 0. That single split drives every modulus question in NDA. Left and right limits diverge at the split point; the function is continuous there but not differentiable.
The principle has been on a tear since 2023 — modulus roughly doubled from about 7 questions per paper-set (2017–22) to about 14 (2023–26), and it has held there for four straight sittings. It now reaches 11 chapters, the widest spread of any principle in the bank: Limits & Continuity, Differentiation, Definite and Indefinite Integration, Apps of Integration, Functions, Quadratic Equations, Sets & Relations, Linear Inequalities, App of Derivatives and Probability. The technique is the same everywhere: split at the zero, handle each piece separately, recombine.
Differentiability is where it is tested hardest. f is differentiable at c only if the left and right derivatives both exist AND agree, and |x| is the canonical counter-example to 'continuous ⇒ differentiable' — at 0 the slopes are −1 and +1. Beware the trap in the other direction: x|x| contains a modulus and IS differentiable at 0, because both one-sided derivatives come out to 0.
Learn to spot the disguise. √(x²) is |x|, not x — so −x/√(x²) is really −x/|x|, a sign function. √(1 − sin 2x) is √((sin x − cos x)²) = |sin x − cos x|, and which branch you take depends entirely on the interval you are given: on (0, π/4) cosine wins, on (π/4, π/2) sine does. The bank sets that same expression twice, once to differentiate and once to integrate, with opposite sign resolutions.
4 worked examples from the bank
Each example demonstrates the principle on a real past-year question. Click to reveal the answer, then the solution.
Example 1 · Limits & Continuity · One-Sided Limits, Greatest Integer, and Absolute Value Limits
Example 2 · Differentiation · Differentiability of Absolute Value, Piecewise, and Greatest Integer Functions
Example 3 · Limits & Continuity · One-Sided Limits, Greatest Integer, and Absolute Value Limits
Example 4 · Limits & Continuity · One-Sided Limits, Greatest Integer, and Absolute Value Limits
Variants to recognise
Same principle, different surfaces. Pattern-match these on test day.
Piecewise definition of |x|
|x| = x for x ≥ 0, −x for x < 0. The split point matters; everything else is algebra.
Left vs right limit at the split
lim x→0⁻ |x|/x = −1; lim x→0⁺ |x|/x = +1. Two-sided limit doesn't exist. NDA exploits this.
|x| is continuous, not differentiable at 0
The graph has a corner. Left derivative = −1, right derivative = +1, so f' undefined at x = 0.
Hidden modulus: √(f²) = |f|
A square root of a perfect square is a modulus, never the bare expression. √(x²) = |x|; √(1 − sin 2x) = |sin x − cos x|. The given interval decides the sign.
The x|x| trap
Contains a modulus yet IS differentiable at 0 — both one-sided derivatives are 0. Presence of |·| is not proof of a corner; always test both sides.
Drill every modulus / absolute value behaviour question
94 questions from the bank — paginated, with cart and Word-export support.
Related principles
Often combined with this one — drill these next if you found the examples above tractable.